How is the integral of int x / ( sinh(x) + cosh(x) ) dx∫xsinh(x)+cosh(x)dx?

2 Answers
Cem Sentin
May 16, 2018

-xe^(-x)-e^(-x)+C−xe−x−e−x+C

Explanation:

int x/(sinhx+coshx)*dx∫xsinhx+coshx⋅dx

=int x/(e^x)*dx∫xex⋅dx

=int xe^(-x)*dx∫xe−x⋅dx

=x*(-e^(-x))-int (-e^(-x))*dxx⋅(−e−x)−∫(−e−x)⋅dx

=-xe^(-x)+int e^(-x)*dx−xe−x+∫e−x⋅dx

=-xe^(-x)-e^(-x)+C−xe−x−e−x+C

Douglas K.
May 16, 2018

Given: int x / ( sinh(x) + cosh(x) ) dx∫xsinh(x)+cosh(x)dx

Multiply the integrand by 1 in the form of (sinh(x)-cosh(x))/(sinh(x)-cosh(x))sinh(x)−cosh(x)sinh(x)−cosh(x):

int x / ( sinh(x) + cosh(x) ) dx = int x / ( sinh(x) + cosh(x) )(sinh(x)-cosh(x))/(sinh(x)-cosh(x)) dx∫xsinh(x)+cosh(x)dx=∫xsinh(x)+cosh(x)sinh(x)−cosh(x)sinh(x)−cosh(x)dx

The denominator is multiplied using the difference of two squares pattern and the numerator is multiplied using the distributive property:

int x / ( sinh(x) + cosh(x) ) dx = int (xsinh(x)-xcosh(x)) / ( sinh^2(x) - cosh^2(x)) dx∫xsinh(x)+cosh(x)dx=∫xsinh(x)−xcosh(x)sinh2(x)−cosh2(x)dx

The identity cosh^2(x)- sinh^2(x)=1cosh2(x)−sinh2(x)=1 tells us that the denominator is -1:

int x / ( sinh(x) + cosh(x) ) dx = int (xsinh(x)-xcosh(x)) / -1 dx∫xsinh(x)+cosh(x)dx=∫xsinh(x)−xcosh(x)−1dx

Eliminate the denominator by changing signs in the numerator:

int x / ( sinh(x) + cosh(x) ) dx = int xcosh(x)-xsinh(x) dx∫xsinh(x)+cosh(x)dx=∫xcosh(x)−xsinh(x)dx

Separate into two integrals:

int x / ( sinh(x) + cosh(x) ) dx = int xcosh(x) dx- int xsinh(x) dx∫xsinh(x)+cosh(x)dx=∫xcosh(x)dx−∫xsinh(x)dx

Both integrals are trivial integrations by parts:

int x / ( sinh(x) + cosh(x) ) dx = (xsinh(x) - cosh(x)) - (xcosh(x)-sinh(x))+C∫xsinh(x)+cosh(x)dx=(xsinh(x)−cosh(x))−(xcosh(x)−sinh(x))+C

Regroup:

int x / ( sinh(x) + cosh(x) ) dx = xsinh(x)- xcosh(x)+ sinh(x)- cosh(x) +C∫xsinh(x)+cosh(x)dx=xsinh(x)−xcosh(x)+sinh(x)−cosh(x)+C

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